We can generalize the problem a bit. Let $g:[0,1]\rightarrow [0,1]$ be a continuous increasing function (in particular it must be invertible). Let the same condition as the original function be satisfied, except that we have $f(r)=[g(r)-1,g(r)+1]\forall r\in [0,1]$. Firstly, note that we can describe $f$ as $f(x)=[l(x),u(x)]$ where $l(x)\leq u(x)\forall x$. We will prove that the only possible $f$ is the following: if $x=m+t$ with $m\in\mathbb{Z}$ and $t\in[0,1)$ with $m$ even, than $l(x)=m-1+g(t)$; if $m$ is odd, $l(x)=m-1+g^{-1}(t)$; in both cases $u(x)=2+l(x)$. We will prove by induction on the intervals $[m,m+1]$ with $m$ integer. Firstly, let $F=\bigcup_{x\in\mathbb{R}}\{x\}\times f(x)$ be the "graph" of $f$, i.e. $(x,y)\in F\iff y\in f(x)$. What condition 1 is saying is that $F$ is symmetric wrt $y=x$, i.e. $(x,y)\in F\iff (x,y)\in F$. The vertical and horizontal segments are sets of the form $F\cap\{x_0\}\times\mathbb{R}$ or $F\cap\mathbb{R}\times\{y_0\}$. The fact that the codomain is $A$ implies that all vertical segments are in fact closed segments (geometrically speaking), so that theor name is appropriate; by symmetry the same is true for horizontal segments. The second condition implies that all horizontal (and thus vertical) segments have length $\leq 2$ (otherwise consider the extremes to get a contradiction). We finally do our "induction", though we only need to do it for positive integers (for negative integers the reasoning is basically the same). As induction hypothesis suppose we have proved that the claim is correct for $x\in[0,m]$ and we want to prove it for $[m,m+1]$. To get $l(x)$ we simply use symmetry. Now, it suffices to prove that $l(m+1+t)=u(m-1+t)\forall t\in[0,1]$, so that by symmetry we can conclude our claimed expression for $u(x)$ in $[m,m+1]$. Condition $2$ says that the horizontal segments of the lines $y=u(m-1+t')$ ($t\in[0,t)$) have length $\leq 2$, so that $l(m+1+t)$ can't be in $u(m-1+[0,t))=[u(m-1),u(m-1+t))=[m,u(m-1+t))$. Furthermore it can't be in $[l(m-1+t),m)$, since the the "lower" boundary of $F$ (i.e. $l(x)$) is also a "right" boundary of $F$, i.e. those points are also the right extremes of their respective horizontal segments. But, we also know by condition $2$ that $f(m-1+t)\cap f(m+1+t)$ is nonempty, implying $l(m+1+t)\leq u(m-1+t)$, which together with the previous conditions implies $l(m+1+t)=u(m-1+t)$ as s wanted.

Let $f(x)=[g(x),h(x)]$. I claim $g$, $h$ is unique and satisfy the following: $g(h(x))=h(g(x))=x, h(x)=g(x)+2=g(x+2), h(x)=1+x^2$ if $0\le x\le 1$ and $2+\sqrt{x-1}$ if $1\le x\le 2$. We first prove a lemma. Lemma: $f(x)$ cannot be a subset of $f(y)$ if $x\ne y$. Proof: Assume for contradiction that's not the case. For now, suppose $x<y$ (the case where $x>y$ is similar). Select $a$ st $x-2<a<\min(x,y-2)$. Then $f(a)\cap f(y)=\emptyset$ because $|a-y|>2$. It follows that $f(a)\cap f(x)=\emptyset$ because $f(x)\subseteq f(y)$. However, $0<x-a<2$, false. We proceed by induction on $n$ to prove that $g$ is increasing, and the following properties hold when $x\in [n,n+1]$: 2. $h(x)=g(x)+2=g(x+2)$ 3. $h(g(x))=g(h(x))=x$ Proof: Base Case: $n=0$. First I prove that $g(1+x)=\sqrt{x}$. Note $1+x\in f(\sqrt{x})$, which implies $\sqrt{x} \in f(1+x)$ so $g(1+x)\le \sqrt{x}$. However, for any $\epsilon>0$, $1+x\notin f(\sqrt{x-\epsilon})$. Therefore, $\sqrt{x-\epsilon} \notin x$ for any $\epsilon>0$ so $g(1+x)=\sqrt{x}$. Similarly, we can prove $h(x-1)=\sqrt{x}$ as well for $0\le x\le 1$. Now I prove $g(2+x)=h(x)$ for $0\le x\le 1$. First, $f(2+x)\cap f(x)\notin \emptyset$. Therefore, $g(2+x)\le h(x)$ and $h(2+x)\ge g(x)$. For any $\epsilon>0$, $f(2+x)\cap f(x-\epsilon)=\emptyset$. Therefore, either $h(2+x)<g(x-\epsilon)$ or $h(x-\epsilon)<g(2+x)$. Since $h(2+x)\ge g(x)>g(x-\epsilon)$ we can see that $h(x-\epsilon)<g(2+x)$. Therefore, $(x-\epsilon)^2+1 < g(2+x) \le x^2+1$ which implies $g(2+x)=h(x)$. Finally I prove $h(1+x)=h(x-1)+2$. By our lemma, we can see $h(1+x)>h(1)=2$. Since $h(1+x)\in f(1+x)$, $x+1\ge g(h(1+x))$ and $1+x\le h(h(1+x))$. Furthermore, since $h(1+x)\notin f(1+x-\epsilon)$, it follows that $1+x-\epsilon \notin f(h(1+x))$, so either $1+x-\epsilon<g(h(1+x))$ or $1+x-\epsilon>h(h(1+x))$. However, $h(h(1+x))\ge 1+x$, so $1+x-\epsilon<g(h(1+x))$ for any $\epsilon$. It follows that $g(h(1+x))=1+x$ for all $0\le x\le 1$. Since $g$ is injective and $g(2+\sqrt{x})=h(\sqrt{x})=1+x$ it follows $h(1+x)=2+\sqrt{x}$. Inductive Step: Assume $g(n-1+x), h(n-1+x), g(n+x)$ follow rules 2 and 3, then I will prove that $h(n+x), g(n+1+x)$ are known to follow these rules as well as well. First, $f(n+1+x)\cap f(n-1-x)\ne \emptyset$ hence it follows that $g(n+1+x) \le h(n-1+x)$ and $h(n+1+x) \ge g(n-1+x)$. However, for any $\epsilon$, we can see $f(n+1+x)\cap f(n-1+x-\epsilon)=\emptyset$ so it follows that $g(n+1+x)>h(n-1+x-\epsilon)$ or $h(n+1+x)<g(n-1+x-\epsilon)$. However, we know $g(n-1+x-\epsilon)<g(n-1+x)\le h(n+1+x)$ so it must follow that $h(n-1+x)=g(n+1+x)$. Similarly we can prove that $h(n+x)=2+h(n-2+x)$. Details omitted, this solves the problem.